Chemistry - Redox Reactions Question with Solution | TestHub

Reduction Half reaction : ${\mathrm{Cr}}_2{{\mathrm{O}}_7}^{2-}+6{\mathrm{e}}^{-}\to 2{\mathrm{Cr}}^{3+}$

Oxidation Half reaction :

${{\mathrm{SO}}_3}^{2-}\to {{\mathrm{SO}}_4}^{2-}+2{\mathrm{e}}^{-}\times 3$

Overall reaction :

${\mathrm{Cr}}_2{{\mathrm{O}}_7}^{2-}+3{{\mathrm{SO}}_3}^{2-}\to 2{\mathrm{Cr}}^{3+}+3{{\mathrm{SO}}_4}^{2-}$

To balance $‘\mathrm{O}’$atoms, adding ${\mathrm{H}}_2\mathrm{O}$ on LHS

${\mathrm{Cr}}_2{{\mathrm{O}}_7}^{2-}+3{{\mathrm{SO}}_3}^{2-}\to 2{\mathrm{Cr}}^{3+}+3{{\mathrm{SO}}_4}^{2-}+4{\mathrm{H}}_2\mathrm{O}$

To balance $‘\mathrm{H}’$ atoms, adding ${\mathrm{H}}^{+}$ on RHS

${\mathrm{Cr}}_2{{\mathrm{O}}_7}^{2-}+3{{\mathrm{SO}}_3}^{2-}+8{\mathrm{H}}^{+}\to 2{\mathrm{Cr}}^{3+}+3{{\mathrm{SO}}_4}^{2-}+4{\mathrm{H}}_2\mathrm{O}$

Therefore, $\mathrm{a}=1, \mathrm{b}=3, \mathrm{c}=8$.